Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int l[maxn], r[maxn];
long long ln[maxn];
double calc(int l, int r, double u, double d) {
  if (r < l) return 0.0;
  double t = ln[r] - (l == 0 ? 0 : ln[l - 1]);
  return t / u * (u - d);
}
double get(double x, double y, double g, double p) {
  return g - (g - x) / y * p;
}
double inter(double l, double r, double x, double y) {
  double tl = max(l, x);
  double tr = min(r, y);
  if (tl > tr) return 0.0;
  return tr - tl;
}
double uni(vector<pair<double, double>> v, double l, double r) {
  vector<pair<double, double>> e;
  for (int i = 0; i < v.size(); ++i) {
    e.emplace_back(v[i].first, 1);
    e.emplace_back(v[i].second, -1);
  }
  sort(e.begin(), e.end());
  double last = 0.0, ans = 0.0;
  int cnt = 0;
  for (int i = 0; i < e.size(); ++i) {
    if (e[i].second == 1) {
      if (cnt == 0) last = e[i].first;
      ++cnt;
    } else {
      --cnt;
      if (cnt == 0) ans += inter(last, e[i].first, l, r);
    }
  }
  return ans;
}
int main() {
  int sy, a, b;
  scanf("%d %d %d", &sy, &a, &b);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d %d", &l[i], &r[i]);
  ln[0] = r[0] - l[0];
  for (int i = 1; i < n; ++i) ln[i] = ln[i - 1] + r[i] - l[i];
  int q;
  scanf("%d", &q);
  int d = -1;
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    int lt = n;
    d = 20;
    while (d--)
      if (lt - (1 << d) >= 0) {
        int to = lt - (1 << d);
        if (l[to] * 1ll * y - (l[to] - x) * 1ll * sy >= a * 1ll * y) lt = to;
      }
    int rg = -1;
    d = 20;
    while (d--)
      if (rg + (1 << d) < n) {
        int to = rg + (1 << d);
        if (r[to] * 1ll * y - (r[to] - x) * 1ll * sy <= b * 1ll * y) rg = to;
      }
    double ans = calc(lt, rg, y, sy);
    vector<pair<double, double>> v;
    if (lt >= 1) {
      double t = get(x, y, r[lt - 1], sy);
      double s = get(x, y, l[lt - 1], sy);
      v.emplace_back(s, t);
    }
    if (rg < n - 1) {
      double s = get(x, y, l[rg + 1], sy);
      double t = get(x, y, r[rg + 1], sy);
      v.emplace_back(s, t);
    }
    ans += uni(v, a, b);
    printf("%.20lf\n", ans);
  }
  return 0;
}